Answer
$y=(\dfrac{2}{5x}+cx^{4})^{-(1/2)}$
Work Step by Step
On comparing the equation $x^2y'+2xy=y^{3}$ with Bernoulli with $n=3$
Consider $p=y^{1-(3)}=y^{-2}$
Then $\dfrac{dy}{dx}=(\dfrac{-1}{2})(p^{-3/2}P\dfrac{dp}{dx}$
or, $\dfrac{dp}{dx}-\dfrac{4p}{x}=-\dfrac{2}{x^2}$
The integrating factor is: $I=e^{\int (\frac{-4}{x})dx}=x^{-4}$
$\int [px^{-4}]'=(-2) \int (x^{-6}) dx$
$\implies p=\dfrac{2}{5x}+cx^{4}$
Hence, $y^{-2}=\dfrac{2}{5x}+cx^{4} \implies y=(\dfrac{2}{5x}+cx^{4})^{-(1/2)}$
